Question

6. An inductive coil is connected in series with resistance of 60-ohm across 230V, 50 Hz ac supply. The voltage
across the coil is 160V and across the resistance is 120 Volts. Calculate (a) the resistance and inductance of
the coil (b) power dissipated in the coil. Also draw the phasor diagram.​

Answer 1

Answer:

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Answer 2

Answer:

Answer:

Given:

Resistance = 6.12 ohms

Inductance = 0.03 H

Potential Difference = 50 V

Frequency = 60 Hz

To calculate:

a ) Inductive Reactance = wL = 2π × f × L

⇒ Inductive Reactance = 2 × 3.14 × 60 Hz × 0.03 H = 11.30 ohms

b ) Impedance ( Z ) = √ ( R² + (wL)² )

⇒ Impedance = √ ( 6.12 )² + ( 11.3 )²

⇒ Impedance = √ ( 37.45 ) + ( 127.69)

⇒ Impedance = √ 165.14 = 12.85 ohms

c ) Current = Potential Difference / Impedance

⇒ Current = 50 V / 12.85 ohms

⇒ Current = 3.89 A

d ) Cos A = R / Z where, A is the phase angle.

⇒ Cos A = 6.12 / 12.85 ≈ 0.5

⇒ A = arccos ( 0.5 ) = 60 degrees = π/3

e ) Power factor = Cos A = 0.5

f  ) Reactive Power = I² ( wL )

⇒ Reactive Power = 3.89 × 3.89 × 11.3 ≈ 171 W

⇒ Active Power = I² R = 3.89 × 3.89 × 6.12 = 92.6 W